Let $a$ be an odd integer $≥3$. It appears that: $$\lim_{n\rightarrow\infty}\frac{1}{2^{n}}\sum_{m=0}^{\left\lfloor n\frac{\ln2}{\ln a}\right\rfloor }\binom{n}{m}=\begin{cases} 1 & \textrm{if }a=3\\ 0 & \textrm{if }a\geq5 \end{cases}$$

Any ideas as to how to prove this? I tried using Stirling's formula, but everything became horrible.

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    $\begingroup$ law of large numbers: the sum of binomials up to $\alpha\cdot n$ is $(1-o(1))2^n$ if $\alpha>1/2$ and $o(2^n)$ if $\alpha<1/2$. $\endgroup$ – Fedor Petrov Aug 28 '19 at 22:53

Fix $0<\alpha<\frac12$ and consider the sum $\sum_{0\le m\le\alpha n} \binom nm$. (For simplicity let's assume $\alpha n$ is an integer.) The ratio of the $m$th term to the $(m+1)$st term in this sum is at most $\alpha/(1-\alpha)$; this means that the sum is bounded by $$ \binom n{\alpha n} \sum_{k=0}^\infty \bigg(\frac\alpha{1-\alpha}\bigg)^k = \binom n{\alpha n} \frac{\alpha-1}{2\alpha-1}, $$ which is $o(2^n)$ by Stirling's formula or probably a simpler argument (an upper bound for the central binomial coefficient will suffice). A similar argument works for the upper tail $\sum_{\beta n\le m\le n} \binom nm$ when $\frac12<\beta<1$.


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